Explanation
So, area of the shaded region $$-\overset{2\sqrt{2}a}{\underset{0}{\int}}x^2/4a\:dx+\overset{2\sqrt{2}a}{\underset{0}{\int}}(2a)\:dx$$ $$2a(2\sqrt{2a})-\dfrac{}{4a}\dfrac{x^3}{3}|_{0}^{2\sqrt{2a}}$$ $$=4\sqrt{2}a^2-\dfrac{(2^{3/2}a)^3}{4a\times 3}$$ $$4\sqrt{2}a^2-\dfrac{2\times2\sqrt{2}a^2}{3}$$ $$=\dfrac{16\sqrt{2}a^2}{3}sq\:units.$$
$$x=y^{2}$$ and the line $$y=x-6$$ area in $$QI$$+ area in QIV $$|\overset {3 }{ \underset { -2}{ \int } } [(y+6)-y^2]dy|$$ $$=|\left | \dfrac{y^2}{2} +6y-y^3/3\right |_{-2}^{3}$$ $$=\left |\dfrac{9}{2} -\dfrac{4}{2}+6(5)-\left [ \dfrac{27}{3} +8/3\right ] \right |$$ $$\left | \dfrac{5}{2}+30-\left [ \dfrac{35}{3} \right ] \right | = \dfrac{15+180-70}{6} = \dfrac{125}{6}sq\:units.$$
Given curves are $$y^2=8x$$ and $$y=2x$$ Let's find out their intersection point
$$\Rightarrow 4x^{2}=8x$$
$$\Rightarrow x^{2}-2x=0$$
$$\Rightarrow x(x-2)=0$$
$$\Rightarrow x=0,2$$
The required area is $$A=\displaystyle \overset{2}{\underset{0}{\int}}(\sqrt{8x}-2x)dx$$ $$=\displaystyle \overset{2}{\underset{0}{\int}}\sqrt{8x}\:dx-\overset{2}{\underset{0}{\int}}2x\:dx$$ $$=\sqrt{8}\times\dfrac{2}{3}\times [x^{3/2}]_{0}^{2}-[x^2]_{0}^{2}$$ $$=\dfrac{2\sqrt{8}}{3}\sqrt{8}-4$$ $$=\dfrac{16}{3}-4=\dfrac{4}{3}\:sq\:units$$
$$I$$: $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{3} \right )^2=1$$ $$Area=\pi \times 2 \times3$$ $$=6\pi sq\:units.$$ $$I$$ $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{2} \right )^2=1$$ $$\Rightarrow x^2+y^2=4$$ $$\pi (2)^2$$ $$=4\pi\:sq\:units.$$
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